91Ó°ÊÓ

A college professor never finishes his lecture before the end of the hour and always finishes his lectures within \(2 \mathrm{~min}\) after the hour. Let \(X=\) the time that elapses between the end of the hour and the end of the lecture and suppose the pdf of \(X\) is $$ f(x)=\left\\{\begin{array}{cl} k x^{2} & 0 \leq x \leq 2 \\ 0 & \text { otherwise } \end{array}\right. $$ a. Find the value of \(k\) and draw the corresponding density curve. [Hint: Total area under the graph of \(f(x)\) is 1.] b. What is the probability that the lecture ends within \(1 \mathrm{~min}\) of the end of the hour? c. What is the probability that the lecture continues beyond the hour for between 60 and \(90 \mathrm{sec}\) ? d. What is the probability that the lecture continues for at least \(90 \mathrm{sec}\) beyond the end of the hour?

Step by step solution

01

Determine the Value of k

The total area under the probability density function (PDF) must equal 1. We integrate the given function \(f(x) = kx^2\) from 0 to 2:\[ \int_{0}^{2} kx^2 \, dx = 1 \]Carrying out the integration:\[ \left[ \frac{kx^3}{3} \right]_{0}^{2} = 1 \]\[ \frac{k (2)^3}{3} - \frac{k (0)^3}{3} = 1 \]\[ \frac{8k}{3} = 1 \]Solving for \(k\), we find:\[ k = \frac{3}{8} \].

02

Draw the Density Curve

The density curve is the graph of the equation \( f(x) = \frac{3}{8} x^2 \) defined over the interval \(0 \leq x \leq 2\). It starts at the origin (0,0) and ends at the point (2,\(\frac{12}{8}\)). It is a curve that grows in a quadratic manner from 0 to 2.

03

Probability Lecture Ends within 1 minute

To find the probability that the lecture ends within 1 minute of the end of the hour, we need to calculate:\[ \int_{0}^{1} \frac{3}{8} x^2 \, dx \]Solving this integral:\[ \frac{3}{8} \left[ \frac{x^3}{3} \right]_{0}^{1} = \frac{3}{8} \times \frac{1}{3} = \frac{1}{8} \].Thus, the probability is \(\frac{1}{8}\).

04

Probability Lecture Ends between 60 and 90 seconds

To find the probability that the lecture continues for between 60 seconds (1 minute) and 90 seconds (1.5 minutes), we calculate:\[ \int_{1}^{1.5} \frac{3}{8} x^2 \, dx \]Solving the integral:\[ \frac{3}{8} \left[ \frac{x^3}{3} \right]_{1}^{1.5} = \frac{3}{8} \left( \frac{(1.5)^3}{3} - \frac{(1)^3}{3} \right) \]\[ = \frac{3}{8} \left( \frac{3.375}{3} - \frac{1}{3} \right) = \frac{3}{8} \times \frac{2.375}{3} = \frac{9.5}{24} = \frac{19}{48} \].The probability is \(\frac{19}{48}\).

05

Probability Lecture Continues for At Least 90 Seconds

The probability that the lecture continues for at least 90 seconds is given by:\[ \int_{1.5}^{2} \frac{3}{8} x^2 \, dx \]Solving this integral:\[ \frac{3}{8} \left[ \frac{x^3}{3} \right]_{1.5}^{2} = \frac{3}{8} \left( \frac{(2)^3}{3} - \frac{(1.5)^3}{3} \right) \]\[ = \frac{3}{8} \left( \frac{8}{3} - \frac{3.375}{3} \right) = \frac{3}{8} \times \frac{4.625}{3} \approx 0.578125 \].Thus, the probability is approximately 0.578.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integration

Integration is a fundamental concept that helps us to find the area under curves, which in probability terms means calculating probabilities when dealing with continuous random variables. In this exercise, we have a probability density function (PDF) defined by a formula involving the variable \( x \). To ensure it is a valid PDF, the total area under the curve must equal 1.
This means integrating the PDF equation over its entire range. The professor’s lecture example provides the PDF as \( f(x) = kx^2 \) and limits between 0 and 2. We used integration:

• Set up the integral: \( \int_{0}^{2} kx^2 \, dx = 1 \).
• Carry out the integration to find the cumulative area.
• The result helps us determine the constant \( k \) since after the integration we solve for \( k \) by setting the integrated value to 1.

This gives us the normalization factor for the PDF ensuring it covers the whole space appropriately.

Continuous Random Variables

Continuous Random Variables are those that can take any value within a given range. In this problem, the variable \( X \) represents the time elapsing after the hour when the professor ends the lecture. Such variables are characterized by PDFs rather than simple counts.
The probability density function \( f(x) \) describes how probabilities are distributed over the range \( 0 \leq x \leq 2 \), representing the additional lecture time:

• If \( X \) is continuous, we cannot simply add up probabilities as we can with discrete variables but must integrate.
• The lecture continues exactly between 0 and 2 minutes, meaning \( f(x) \) is 0 outside this interval.
• By integrating \( f(x) \) over any sub-range, we find the probability for that specific time period.

This approach highlights a difference from discrete variables, emphasizing the role of integration in defining and calculating probabilities.

Probability Calculations

When dealing with continuous random variables, calculating probabilities involves finding the area under the curve of the PDF for a specific interval. This requires setting up an integral over the interval of interest. Here’s how we handled it in our exercise:

• For the lecture ending within 1 minute, we computed \( \int_0^1 \frac{3}{8}x^2 \, dx \) and found it to be \( \frac{1}{8} \).
• To find the probability for the lecture continuing between 60 to 90 seconds, we evaluated \( \int_1^{1.5} \frac{3}{8}x^2 \, dx \) yielding \( \frac{19}{48} \).
• Finally, for at least 90 seconds, \( \int_{1.5}^{2} \frac{3}{8}x^2 \, dx \) resulted in approximately 0.578.

These calculations demonstrate how integration can transform a PDF into meaningful probabilities for continuous random events. It’s a vital tool for making sense of data distributions in practical terms, reflecting real-life scenarios like lecture timings.